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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1 TZ2TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Q1 Q2 Q3 Q4 Q5

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

14 marks

Compound interest, savings plan and an arithmetic sequence

Number and algebra

(a)

3 marks

Interest is added monthly, so use $12$ periods a year for $3$ years: $36$ periods in all.
Enter it in the finance app (N = 36, I% = 2.3, PV = −14500, PMT = 0, P/Y = C/Y = 12) or use the compound interest formula.
Round to two decimal places, as the paper asks.

F V = 14500 (1 + \frac{2.3}{100 \times 12})^{36} F V = 15534.80 GBP (GDC)

Mentor tip

Keep the signs consistent in the finance app: money you pay in (PV) is negative, so FV comes out positive.

(b)

3 marks

Your answer to (a) becomes the new starting value, and the monthly deposit goes in as PMT.
Two years of monthly payments means $24$ periods. Payments are at the end of each month, so leave the app in its default (end) mode.

N = 24, I % = 2.3, P V = - 15534.80, P M T = - 250, P / Y = C / Y = 12 F V = 22399.50 GBP (GDC)

Mentor tip

PV and PMT are both money going into the account, so give them the same sign.

(c)

5 marks

(i) The common difference is $26616 - 28492 = - 1876$ . Write $u_{n} = u_{1} + (n - 1) d$ and simplify to the form $a + bn$ .

(ii) Put $n = 7$ into your formula.

(iii) Give one clear reason the straight-line model breaks down, for example that it gives a negative price for cars older than about $16$ years, while a real car always keeps some scrap value. Another good answer: cars of the same age can differ in price because of mileage or service history, which the formula ignores.

u_{n} = 28492 + (n - 1) (- 1876) u_{n} = 30368 - 1876 n u_{7} = 30368 - 1876 (7) = 17236 GBP

Mentor tip

For (iii), say how the model fails for some values of $n$ . Just writing “condition of the car” is too vague, because it does not explain why the price would change for the same age.

(d)

3 marks

Newer cars cost more, so the youngest car she can afford is the first one priced at or below her balance of $22399.50$ .
Solve $u_{n} \leq 22399.50$ : this gives $n \geq 4.25$ , so the youngest affordable car is $5$ years old.
Subtract its price from the balance.

30368 - 1876 n \leq 22399.50 ⟹ n \geq 4.248 \dots u_{5} = 30368 - 1876 (5) = 20988 22399.50 - 20988 = 1411.50 GBP

Mentor tip

Age must be a whole number, so round $n$ up, not to the nearest integer: a $4$ -year-old car costs $22864$ , which is more than she has.

Question 2

18 marks

Summary statistics, a t-test and linear regression

Statistics and probability

(a)

4 marks

Enter each row of data as a list and run one-variable statistics on it.
Read off the mean and the standard deviation $σ_{x}$ for each group.

(i) \overset{x}{ˉ}_{A} = 122.25 \approx 122, σ_{A} = 8.56956 \dots \approx 8.57 (GDC) (ii) \overset{x}{ˉ}_{N} = 131.375 \approx 131, σ_{N} = 8.57230 \dots \approx 8.57 (GDC)

Mentor tip

The GDC shows two standard deviations. Use $σ_{x}$ , the one for the data set itself, not $s_{x}$ (about $9.16$ ).

(b)

2 marks

Compare the centres: the non-active group has the higher mean heart rate.
Compare the spreads: the standard deviations are almost identical, so both groups are equally spread about their means.

\overset{x}{ˉ}_{N} > \overset{x}{ˉ}_{A}, σ_{A} \approx σ_{N}

Mentor tip

“Compare and contrast” needs one similarity and one difference. Refer to the context (heart rate), not just the numbers.

(c)

2 marks

(i) The null hypothesis says there is no difference between the population means.

(ii) The alternative matches the belief being tested: the active mean is lower, so the test is one-tailed.

(i) H_{0} : μ_{A} = μ_{N} (ii) H_{1} : μ_{A} < μ_{N}

Mentor tip

Hypotheses are about population means, so use $μ$ , never $\overset{x}{ˉ}$ .

(d)

2 marks

Run a two-sample t-test on the lists, with the active list first, the alternative set to $μ_{1} < μ_{2}$ and pooled variance chosen.

t = - 1.99 \dots p = 0.0331372 \dots \approx 0.0331 (GDC)

Mentor tip

If your $p$ -value is about $0.967$ , you set the inequality the wrong way round. Check the list order and the direction.

(e)

2 marks

Compare the $p$ -value with the $5%$ significance level.
It is smaller, so reject $H_{0}$ . There is enough evidence that the mean heart rate of active people is lower than that of non-active people.

0.0331 < 0.05 ⟹ reject H_{0}

Mentor tip

“In context” means finishing with a sentence about heart rates, not just “reject $H_{0}$ ”.

(f)

3 marks

Enter the ages as $x$ and the heart rates as $y$ , then run linear regression ( $a x + b$ ).
Read off the gradient, the intercept and $r$ .

(i) y = 1.375 x + 87.1875 \approx 1.38 x + 87.2 (GDC) (ii) r = 0.849029 \dots \approx 0.849

Mentor tip

Make sure $x$ is the list of ages. Swapping the lists gives the regression of $x$ on $y$ , a different line.

(g)

3 marks

(i) Substitute $x = 45$ into the full-accuracy regression equation.

(ii) Her age is well outside the ages in the data ( $18$ to $33$ ), so the prediction is an extrapolation and may not be reliable.

y = 1.375 (45) + 87.1875 = 149.0625 y \approx 149 beats per minute

Mentor tip

Name the reason exactly: “extrapolation, because 45 is outside the data range” is much stronger than “the model may be wrong”.

Question 3

12 marks

A sine model for body temperature

Functions

(a)

2 marks

For $sin (b t)$ the period is $\frac{2 π}{b}$ in radians (or $\frac{360}{b}$ in degrees). Set it equal to $24$ and solve.

\frac{2 π}{b} = 24 ⟹ b = \frac{π}{12} \approx 0.262 (in degrees: b = 15)

Mentor tip

Whichever you choose, keep your calculator in the matching mode for the rest of the question.

(b)

6 marks

(i) At $18 : 00$ , $t = 18$ and $b t = \frac{3 π}{2}$ (or $270°$ ), where sine equals $- 1$ .

(ii) At $13 : 00$ , $t = 13$ . Work out $sin \frac{13 π}{12}$ and substitute.

(iii) Solve the two equations together, either by subtracting or with the simultaneous equation solver.

36.9 = a sin (\frac{π}{12} \times 18) + c = a sin \frac{3 π}{2} + c = - a + c 36.6 = a sin (\frac{13 π}{12}) + c = - 0.258819 \dots a + c - 0.741180 \dots a = 0.3 a = - 0.404759 \dots \approx - 0.405 c = 36.9 + a = 36.4952 \dots \approx 36.5

Mentor tip

A negative $a$ is fine here: it flips the sine curve, so the temperature falls first after midnight. Do not force it to be positive.

(c)

2 marks

Put $t = 8$ into the model, using the unrounded $a$ and $c$ .
Compare the model's value with the reading of $36.2$ .

T (8) = - 0.404759 \dots sin (\frac{8 π}{12}) + 36.4952 \dots = 36.1447 \dots 36.2 > 36.1 ⟹ the point lies above the curve

Mentor tip

Rounding $a$ and $c$ to $3$ s.f. first can shift $T (8)$ enough to matter. Store the full values in your GDC.

(d)

2 marks

Because $a < 0$ , the curve starts by falling from its midline at $t = 0$ to a minimum at $t = 6$ , rises to a maximum at $t = 18$ , then falls again to $t = 24$ .
So the temperature is decreasing over two stretches of the day.

0 < t < 6 (00:00 to 06:00) 18 < t < 24 (18:00 to 24:00)

Mentor tip

Sketch or graph the model before answering. Both intervals are needed, and the one late in the evening is easy to forget.

Question 4

17 marks

Wind turbine power, bounds and a polynomial model

Number and algebra
Calculus
Functions

(a)

3 marks

Substitute $r = 40$ and $v = 10$ into the power formula.
Write the result in standard form, correct to $3$ s.f.

P = 0.245 π (40)^{2} (10)^{3} = 1231504.3 \dots P \approx 1.23 \times 1 0^{6}

Mentor tip

Standard form means $a \times 1 0^{k}$ with $1 \leq a < 10$ . Calculator notation such as 1.23E6 is not a final answer.

(b)

5 marks

Speeds in the table are rounded to $1$ d.p., so category $6$ really covers $10.75 \leq v < 13.85$ .
(i) The least power comes from the lower bound $v = 10.75$ .
(ii) The greatest power comes from the upper bound $v = 13.85$ . Subtract the smallest power from it.

P_{m i n} = 0.245 π (40)^{2} (10.75)^{3} = 1529893.9 \dots \approx 1.53 \times 1 0^{6} P_{m a x} = 0.245 π (40)^{2} (13.85)^{3} = 3271788.7 \dots P_{m a x} - P_{m i n} = 1741894.8 \dots \approx 1.74 \times 1 0^{6}

Mentor tip

The trap is using $10.8$ and $13.8$ from the table. Always ask what range of true values rounds to the numbers shown.

(c)

6 marks

(i) Differentiate each term with the power rule; the constant disappears.

(ii) Solve $\frac{d P}{d v} = 0$ on the GDC (polynomial solver or graph) and keep the root inside $3.35 \leq v < 17.15$ .

(iii) The graph of $P$ has a maximum there, so this is the wind speed at which the second turbine produces its greatest power.

\frac{d P}{d v} = 17.92 v^{3} - 708 v^{2} + 7140 v - 8740 17.92 v^{3} - 708 v^{2} + 7140 v - 8740 = 0 v = 14.7728 \dots \approx 14.8 m s^{- 1} (GDC)

Mentor tip

The cubic has other roots outside the operating range, so check the domain. For (iii), say “maximum power”, not just “turning point”.

(d)

3 marks

Graph $P$ and the line $P = 90000$ on the domain $3.35 \leq v < 17.15$ and find where they cross.
The curve crosses once, at $v \approx 11.46$ , and stays above $90000$ up to the edge of the operating range at $17.15$ .

4.48 v^{4} - 236 v^{3} + 3570 v^{2} - 8740 v - 735 = 90000 v = 11.4550 \dots (GDC) 11.46 \leq v < 17.15

Mentor tip

The upper end comes from the domain, not from a second intersection, so keep the strict “ $<$ ” at $17.15$ .

Question 5

19 marks

Making a cone from a sector and maximising its volume

Geometry and trigonometry
Calculus

(a)

4 marks

(i) Arc length is the fraction $\frac{θ}{360}$ of the full circumference $2 π \times 9 = 18 π$ .

(ii) Take the removed arc away from the whole circumference, or use the angle $360 - θ$ directly.

x = \frac{θ}{360} \times 18 π = \frac{π θ}{20} remaining arc = 18 π - \frac{π θ}{20}

Mentor tip

The angle is in degrees, so use $\frac{θ}{360}$ , not the radian formula $r θ$ .

(b)

2 marks

When the sector is rolled up, the remaining arc becomes the circle round the base of the cone.
Set it equal to $2 π r$ and divide by $2 π$ .

2 π r = 18 π - \frac{π θ}{20} r = 9 - \frac{θ}{40}

Mentor tip

On a “show that”, write the linking statement in words (arc length equals base circumference) before the algebra.

(c)

3 marks

The radius, height and slant height form a right-angled triangle, so $r^{2} + h^{2} = 9^{2}$ .
Make $h$ the subject, then substitute $h$ and $r = 9 - \frac{θ}{40}$ into $V = \frac{1}{3} π r^{2} h$ .

h = 81 - r^{2} V = \frac{1}{3} π r^{2} 81 - r^{2} V = \frac{1}{3} π (9 - \frac{θ}{40})^{2} 81 - (9 - \frac{θ}{40})^{2}

Mentor tip

Show the Pythagoras step and the substitution line; jumping straight to the final expression does not show anything.

(d)

5 marks

(i) Graph $V$ against $θ$ for $0 < θ < 360$ and use the maximum feature.

(ii) The $y$ -value of that point is the largest volume.

(iii) Put the full-accuracy $θ$ into $r = 9 - \frac{θ}{40}$ .

θ = 66.0612 \dots \approx 66.1 (GDC) V_{m a x} = 293.835 \dots \approx 294 cm^{3} r = 9 - \frac{66.0612 \dots}{40} = 7.34846 \dots \approx 7.35 cm

Mentor tip

Set a sensible window (for example $θ$ from $0$ to $360$ , $V$ from $0$ to $350$ ) or the maximum may be off screen.

(e)

5 marks

The hemisphere sits on the cone's opening, so it has the same radius, $r = 7.348 \dots$ cm.
Its volume is half a sphere: $\frac{2}{3} π r^{3}$ .
Add the maximum cone volume and compare the total with $1000$ cm $^{3}$ .

V_{hemisphere} = \frac{2}{3} π (7.34846 \dots)^{3} = 831.089 \dots V_{total} = 831.089 \dots + 293.835 \dots = 1124.92 \dots cm^{3} 1124.92 > 1000 ⟹ the claim is true

Mentor tip

End with a clear verdict that refers back to $1000$ cm $^{3}$ (one litre). A number with no conclusion leaves the question unanswered.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra22 marks · 28%
Q1, Q4
Statistics and probability18 marks · 23%
Q2
Functions15 marks · 19%
Q3, Q4
Geometry and trigonometry14 marks · 18%
Q5
Calculus11 marks · 14%
Q4, Q5

Mentor insights

Traps to watch for

01
Rounding the car's age the wrong way · Q1
Solving the inequality gives n ≈ 4.25. Rounding to 4 picks a car she cannot afford; age must round up to 5.
02
Picking the wrong standard deviation · Q2
The GDC shows σx and sx. For describing these data sets, use σx (8.57), not sx (about 9.16).
03
Forgetting that a is negative · Q3
With a < 0 the sine curve is upside down, so the temperature falls from midnight to 06:00 and again after 18:00.
04
Bounds from rounded speeds · Q4
Category 6 really runs from 10.75 to 13.85 m/s. Using the printed 10.8 and 13.8 gives the wrong powers.
05
Hemisphere volume · Q5
A hemisphere is half a sphere, (2/3)πr³, using the cone's radius. Using the full sphere or the slant height of 9 cm overstates the total.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1 TZ2TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Q1 Q2 Q3 Q4 Q5

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

14 marks

Compound interest, savings plan and an arithmetic sequence

Number and algebra

(a)

3 marks

Interest is added monthly, so use $12$ periods a year for $3$ years: $36$ periods in all.
Enter it in the finance app (N = 36, I% = 2.3, PV = −14500, PMT = 0, P/Y = C/Y = 12) or use the compound interest formula.
Round to two decimal places, as the paper asks.

F V = 14500 (1 + \frac{2.3}{100 \times 12})^{36} F V = 15534.80 GBP (GDC)

Mentor tip

Keep the signs consistent in the finance app: money you pay in (PV) is negative, so FV comes out positive.

(b)

3 marks

Your answer to (a) becomes the new starting value, and the monthly deposit goes in as PMT.
Two years of monthly payments means $24$ periods. Payments are at the end of each month, so leave the app in its default (end) mode.

N = 24, I % = 2.3, P V = - 15534.80, P M T = - 250, P / Y = C / Y = 12 F V = 22399.50 GBP (GDC)

Mentor tip

PV and PMT are both money going into the account, so give them the same sign.

(c)

5 marks

(i) The common difference is $26616 - 28492 = - 1876$ . Write $u_{n} = u_{1} + (n - 1) d$ and simplify to the form $a + bn$ .

(ii) Put $n = 7$ into your formula.

u_{n} = 28492 + (n - 1) (- 1876) u_{n} = 30368 - 1876 n u_{7} = 30368 - 1876 (7) = 17236 GBP

Mentor tip

For (iii), say how the model fails for some values of $n$ . Just writing “condition of the car” is too vague, because it does not explain why the price would change for the same age.

(d)

3 marks

Newer cars cost more, so the youngest car she can afford is the first one priced at or below her balance of $22399.50$ .
Solve $u_{n} \leq 22399.50$ : this gives $n \geq 4.25$ , so the youngest affordable car is $5$ years old.
Subtract its price from the balance.

30368 - 1876 n \leq 22399.50 ⟹ n \geq 4.248 \dots u_{5} = 30368 - 1876 (5) = 20988 22399.50 - 20988 = 1411.50 GBP

Mentor tip

Age must be a whole number, so round $n$ up, not to the nearest integer: a $4$ -year-old car costs $22864$ , which is more than she has.

Question 2

18 marks

Summary statistics, a t-test and linear regression

Statistics and probability

(a)

4 marks

Enter each row of data as a list and run one-variable statistics on it.
Read off the mean and the standard deviation $σ_{x}$ for each group.

(i) \overset{x}{ˉ}_{A} = 122.25 \approx 122, σ_{A} = 8.56956 \dots \approx 8.57 (GDC) (ii) \overset{x}{ˉ}_{N} = 131.375 \approx 131, σ_{N} = 8.57230 \dots \approx 8.57 (GDC)

Mentor tip

The GDC shows two standard deviations. Use $σ_{x}$ , the one for the data set itself, not $s_{x}$ (about $9.16$ ).

(b)

2 marks

Compare the centres: the non-active group has the higher mean heart rate.
Compare the spreads: the standard deviations are almost identical, so both groups are equally spread about their means.

\overset{x}{ˉ}_{N} > \overset{x}{ˉ}_{A}, σ_{A} \approx σ_{N}

Mentor tip

“Compare and contrast” needs one similarity and one difference. Refer to the context (heart rate), not just the numbers.

(c)

2 marks

(i) The null hypothesis says there is no difference between the population means.

(ii) The alternative matches the belief being tested: the active mean is lower, so the test is one-tailed.

(i) H_{0} : μ_{A} = μ_{N} (ii) H_{1} : μ_{A} < μ_{N}

Mentor tip

Hypotheses are about population means, so use $μ$ , never $\overset{x}{ˉ}$ .

(d)

2 marks

Run a two-sample t-test on the lists, with the active list first, the alternative set to $μ_{1} < μ_{2}$ and pooled variance chosen.

t = - 1.99 \dots p = 0.0331372 \dots \approx 0.0331 (GDC)

Mentor tip

If your $p$ -value is about $0.967$ , you set the inequality the wrong way round. Check the list order and the direction.

(e)

2 marks

Compare the $p$ -value with the $5%$ significance level.
It is smaller, so reject $H_{0}$ . There is enough evidence that the mean heart rate of active people is lower than that of non-active people.

0.0331 < 0.05 ⟹ reject H_{0}

Mentor tip

“In context” means finishing with a sentence about heart rates, not just “reject $H_{0}$ ”.

(f)

3 marks

Enter the ages as $x$ and the heart rates as $y$ , then run linear regression ( $a x + b$ ).
Read off the gradient, the intercept and $r$ .

(i) y = 1.375 x + 87.1875 \approx 1.38 x + 87.2 (GDC) (ii) r = 0.849029 \dots \approx 0.849

Mentor tip

Make sure $x$ is the list of ages. Swapping the lists gives the regression of $x$ on $y$ , a different line.

(g)

3 marks

(i) Substitute $x = 45$ into the full-accuracy regression equation.

(ii) Her age is well outside the ages in the data ( $18$ to $33$ ), so the prediction is an extrapolation and may not be reliable.

y = 1.375 (45) + 87.1875 = 149.0625 y \approx 149 beats per minute

Mentor tip

Name the reason exactly: “extrapolation, because 45 is outside the data range” is much stronger than “the model may be wrong”.

Question 3

12 marks

A sine model for body temperature

Functions

(a)

2 marks

For $sin (b t)$ the period is $\frac{2 π}{b}$ in radians (or $\frac{360}{b}$ in degrees). Set it equal to $24$ and solve.

\frac{2 π}{b} = 24 ⟹ b = \frac{π}{12} \approx 0.262 (in degrees: b = 15)

Mentor tip

Whichever you choose, keep your calculator in the matching mode for the rest of the question.

(b)

6 marks

(i) At $18 : 00$ , $t = 18$ and $b t = \frac{3 π}{2}$ (or $270°$ ), where sine equals $- 1$ .

(ii) At $13 : 00$ , $t = 13$ . Work out $sin \frac{13 π}{12}$ and substitute.

(iii) Solve the two equations together, either by subtracting or with the simultaneous equation solver.

36.9 = a sin (\frac{π}{12} \times 18) + c = a sin \frac{3 π}{2} + c = - a + c 36.6 = a sin (\frac{13 π}{12}) + c = - 0.258819 \dots a + c - 0.741180 \dots a = 0.3 a = - 0.404759 \dots \approx - 0.405 c = 36.9 + a = 36.4952 \dots \approx 36.5

Mentor tip

A negative $a$ is fine here: it flips the sine curve, so the temperature falls first after midnight. Do not force it to be positive.

(c)

2 marks

Put $t = 8$ into the model, using the unrounded $a$ and $c$ .
Compare the model's value with the reading of $36.2$ .

T (8) = - 0.404759 \dots sin (\frac{8 π}{12}) + 36.4952 \dots = 36.1447 \dots 36.2 > 36.1 ⟹ the point lies above the curve

Mentor tip

Rounding $a$ and $c$ to $3$ s.f. first can shift $T (8)$ enough to matter. Store the full values in your GDC.

(d)

2 marks

Because $a < 0$ , the curve starts by falling from its midline at $t = 0$ to a minimum at $t = 6$ , rises to a maximum at $t = 18$ , then falls again to $t = 24$ .
So the temperature is decreasing over two stretches of the day.

0 < t < 6 (00:00 to 06:00) 18 < t < 24 (18:00 to 24:00)

Mentor tip

Sketch or graph the model before answering. Both intervals are needed, and the one late in the evening is easy to forget.

Question 4

17 marks

Wind turbine power, bounds and a polynomial model

Number and algebra
Calculus
Functions

(a)

3 marks

Substitute $r = 40$ and $v = 10$ into the power formula.
Write the result in standard form, correct to $3$ s.f.

P = 0.245 π (40)^{2} (10)^{3} = 1231504.3 \dots P \approx 1.23 \times 1 0^{6}

Mentor tip

Standard form means $a \times 1 0^{k}$ with $1 \leq a < 10$ . Calculator notation such as 1.23E6 is not a final answer.

(b)

5 marks

Speeds in the table are rounded to $1$ d.p., so category $6$ really covers $10.75 \leq v < 13.85$ .
(i) The least power comes from the lower bound $v = 10.75$ .
(ii) The greatest power comes from the upper bound $v = 13.85$ . Subtract the smallest power from it.

P_{m i n} = 0.245 π (40)^{2} (10.75)^{3} = 1529893.9 \dots \approx 1.53 \times 1 0^{6} P_{m a x} = 0.245 π (40)^{2} (13.85)^{3} = 3271788.7 \dots P_{m a x} - P_{m i n} = 1741894.8 \dots \approx 1.74 \times 1 0^{6}

Mentor tip

The trap is using $10.8$ and $13.8$ from the table. Always ask what range of true values rounds to the numbers shown.

(c)

6 marks

(i) Differentiate each term with the power rule; the constant disappears.

(ii) Solve $\frac{d P}{d v} = 0$ on the GDC (polynomial solver or graph) and keep the root inside $3.35 \leq v < 17.15$ .

(iii) The graph of $P$ has a maximum there, so this is the wind speed at which the second turbine produces its greatest power.

\frac{d P}{d v} = 17.92 v^{3} - 708 v^{2} + 7140 v - 8740 17.92 v^{3} - 708 v^{2} + 7140 v - 8740 = 0 v = 14.7728 \dots \approx 14.8 m s^{- 1} (GDC)

Mentor tip

The cubic has other roots outside the operating range, so check the domain. For (iii), say “maximum power”, not just “turning point”.

(d)

3 marks

Graph $P$ and the line $P = 90000$ on the domain $3.35 \leq v < 17.15$ and find where they cross.
The curve crosses once, at $v \approx 11.46$ , and stays above $90000$ up to the edge of the operating range at $17.15$ .

4.48 v^{4} - 236 v^{3} + 3570 v^{2} - 8740 v - 735 = 90000 v = 11.4550 \dots (GDC) 11.46 \leq v < 17.15

Mentor tip

The upper end comes from the domain, not from a second intersection, so keep the strict “ $<$ ” at $17.15$ .

Question 5

19 marks

Making a cone from a sector and maximising its volume

Geometry and trigonometry
Calculus

(a)

4 marks

(i) Arc length is the fraction $\frac{θ}{360}$ of the full circumference $2 π \times 9 = 18 π$ .

(ii) Take the removed arc away from the whole circumference, or use the angle $360 - θ$ directly.

x = \frac{θ}{360} \times 18 π = \frac{π θ}{20} remaining arc = 18 π - \frac{π θ}{20}

Mentor tip

The angle is in degrees, so use $\frac{θ}{360}$ , not the radian formula $r θ$ .

(b)

2 marks

When the sector is rolled up, the remaining arc becomes the circle round the base of the cone.
Set it equal to $2 π r$ and divide by $2 π$ .

2 π r = 18 π - \frac{π θ}{20} r = 9 - \frac{θ}{40}

Mentor tip

On a “show that”, write the linking statement in words (arc length equals base circumference) before the algebra.

(c)

3 marks

The radius, height and slant height form a right-angled triangle, so $r^{2} + h^{2} = 9^{2}$ .
Make $h$ the subject, then substitute $h$ and $r = 9 - \frac{θ}{40}$ into $V = \frac{1}{3} π r^{2} h$ .

h = 81 - r^{2} V = \frac{1}{3} π r^{2} 81 - r^{2} V = \frac{1}{3} π (9 - \frac{θ}{40})^{2} 81 - (9 - \frac{θ}{40})^{2}

Mentor tip

Show the Pythagoras step and the substitution line; jumping straight to the final expression does not show anything.

(d)

5 marks

(i) Graph $V$ against $θ$ for $0 < θ < 360$ and use the maximum feature.

(ii) The $y$ -value of that point is the largest volume.

(iii) Put the full-accuracy $θ$ into $r = 9 - \frac{θ}{40}$ .

θ = 66.0612 \dots \approx 66.1 (GDC) V_{m a x} = 293.835 \dots \approx 294 cm^{3} r = 9 - \frac{66.0612 \dots}{40} = 7.34846 \dots \approx 7.35 cm

Mentor tip

Set a sensible window (for example $θ$ from $0$ to $360$ , $V$ from $0$ to $350$ ) or the maximum may be off screen.

(e)

5 marks

The hemisphere sits on the cone's opening, so it has the same radius, $r = 7.348 \dots$ cm.
Its volume is half a sphere: $\frac{2}{3} π r^{3}$ .
Add the maximum cone volume and compare the total with $1000$ cm $^{3}$ .

V_{hemisphere} = \frac{2}{3} π (7.34846 \dots)^{3} = 831.089 \dots V_{total} = 831.089 \dots + 293.835 \dots = 1124.92 \dots cm^{3} 1124.92 > 1000 ⟹ the claim is true

Mentor tip

End with a clear verdict that refers back to $1000$ cm $^{3}$ (one litre). A number with no conclusion leaves the question unanswered.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra22 marks · 28%
Q1, Q4
Statistics and probability18 marks · 23%
Q2
Functions15 marks · 19%
Q3, Q4
Geometry and trigonometry14 marks · 18%
Q5
Calculus11 marks · 14%
Q4, Q5

Mentor insights

Traps to watch for

01
Rounding the car's age the wrong way · Q1
Solving the inequality gives n ≈ 4.25. Rounding to 4 picks a car she cannot afford; age must round up to 5.
02
Picking the wrong standard deviation · Q2
The GDC shows σx and sx. For describing these data sets, use σx (8.57), not sx (about 9.16).
03
Forgetting that a is negative · Q3
With a < 0 the sine curve is upside down, so the temperature falls from midnight to 06:00 and again after 18:00.
04
Bounds from rounded speeds · Q4
Category 6 really runs from 10.75 to 13.85 m/s. Using the printed 10.8 and 13.8 gives the wrong powers.
05
Hemisphere volume · Q5
A hemisphere is half a sphere, (2/3)πr³, using the cone's radius. Using the full sphere or the slant height of 9 cm overstates the total.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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